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using cosh(x) ? - Printable Version

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using cosh(x) ? - tommy1729 - 11/19/2011

no , im not kidding , apart from my sinh method im considering a cosh method.

although it might be more complicated or speculative.

it might be insightfull to plot things.

consider

exp(1/e)<b<sqrt(e)

f(x) - f^[-1](x) = 2*cosh(ln(b)*x)

for variable x and fixed b ( base ).

i bet you can find more functional equations from this one.

for instance to express f^[-1](x) in terms of f(x) and elementary functions.

the solution is unique for R-> R.

replace x by f(x) and let g(x) be the super of f(x).

then we get a difference equation for g(x).

we solve that difference equation , now we go from g(x) to f(x) or we find the taylor for f(x) from the equation ( or both , in programming ).

similar to the sinh method : since f is close to 2*cosh(ln(b)*x)) and 2*cosh(ln(b)*x) is close to b^x , we have a new method.


RE: using cosh(x) ? - tommy1729 - 11/20/2011

also , we finally have boundaries.

for large x , f^[1/2](x) + f^[1/2](ln_b(x)) is a good bound for tet(b,1/2,x).


RE: using cosh(x) ? - tommy1729 - 11/20/2011

hmm

i was thinking about

f(x) - f[-1](x) = 2*sinh(ln(b)*x).

for approximating (b^x)^[theta] for bases b with eta < b < sqrt(e) and x > e^e.




RE: using cosh(x) ? - tommy1729 - 11/20/2011

a possible improvement would be

f(x) - f^[-1](x) = 2*sinh(ln(b)*x) + (e-2)*x/e.

f(0) = 0

tommy1729


RE: using cosh(x) ? - sheldonison - 11/20/2011

(11/19/2011, 11:56 PM)tommy1729 Wrote: no , im not kidding , apart from my sinh method im considering a cosh method.
What is the fixed point for this method? The fixed point for 2sinh was zero.



RE: using cosh(x) ? - tommy1729 - 11/20/2011

(11/20/2011, 03:18 PM)sheldonison Wrote:
(11/19/2011, 11:56 PM)tommy1729 Wrote: no , im not kidding , apart from my sinh method im considering a cosh method.
What is the fixed point for this method? The fixed point for 2sinh was zero.

the fixed point is pi/2 i for the cosh method.

for the ' new ' sinh method it is also 0.